A fire hydrant delivers water of density - at a volume rate l- The water travels vertically upwards through the hydrant and then does 900 turn to emerge horizontally at speed - The pipe and nozzle have uniform cross-section throughout- The force exerted by water on the corner of the hydrant is

In time Δ t, momentum of water entering the hydrant nbsp; P_1=(ρ L Δ t) vĵ Momentum of water while leaving the hydrant in time Δ t is P_2=(ρ L Δ t) v( î) Ch ...

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Byju's Answer

Standard XII

Chemistry

Longest Chain Rule

A fire hydran...

Question

A fire hydrant delivers water of density ρ at a volume rate l. The water travels vertically upwards through the hydrant and then does 90⁰ turn to emerge horizontally at speed υ. The pipe and nozzle have uniform cross-section throughout. The force exerted by water on the corner of the hydrant is

Zero

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$ρ v l$

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$2 ρ v l$

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$\sqrt{2} ρ v l$

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Solution

The correct option is D
$\sqrt{2} ρ v l$

In time $Δ t$ , momentum of water entering the hydrant
$P_{1} = (ρ L Δ t) v^j$

Momentum of water while leaving the hydrant in time $Δ t$ is
$P_{2} = (ρ L Δ t) v (-^i)$
Change in momentum in time $Δ t$ is
$Δ P = P_{2} - P_{1} = ρ L t (-^i -^j)$

$| Δ P | = ρ L Δ t v \sqrt{(- 1)^{2} + (- 1)^{2}}$
$= \sqrt{2} ρ L Δ t v$

$Force exerted by water, F = F = \frac{| Δ P |}{Δ t} = \sqrt{2} ρ L v$

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A fire hydrant delivers water of density ρ at a volume rate l. The water travels vertically upwards through the hydrant and then does 900 turn to emerge horizontally at speed υ. The pipe and nozzle have uniform cross-section throughout. The force exerted by water on the corner of the hydrant is

A fire hydrant delivers water of density ρ at a volume rate l. The water travels vertically upwards through the hydrant and then does 90⁰ turn to emerge horizontally at speed υ. The pipe and nozzle have uniform cross-section throughout. The force exerted by water on the corner of the hydrant is